Question

A wire of length 20 units is divided into two parts such that the product of one part and the cube of the other part is maximum. The product of these parts is:

• A. \(5\)
• B. \(75\)
• C. \(15\)
• D. \(70\)

Hint:

Use derivatives to find maximum or minimum values in optimization problems.

Answer 1

The Correct Option is B

Let \( x \) be the one part and \( y \) be the other part. We have: \[ x + y = 20 \implies y = 20 - x \] As per the given condition, we write: \[ f(x) = (20 - x)x^3 = 20x^3 - x^4 \] Differentiating \( f(x) \) with respect to \( x \): \[ f'(x) = 60x^2 - 4x^3 \] Setting \( f'(x) = 0 \), we solve: \[ 4x^2(15 - x) = 0 \implies x = 0, 15 \] Now, compute the second derivative: \[ f''(x) = 120x - 12x^2 \] Evaluate \( f''(x) \) at \( x = 15 \): \[ f''(15) = (120)(15) - (12)(15)^2 = 1800 - 2700 = -900 < 0 \] Thus, \( f(x) \) is maximum when \( x = 15 \). \[ \therefore f(x) \text{ is maximum when } x = 15. \] From \( y = 20 - x \), we get: \[ y = 20 - 15 = 5 \] The product \( xy \) is: \[ xy = (15)(5) = 75 \]
Final Answer: \( xy = 75 \).